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Romashka-Z-Leto [24]
3 years ago
15

Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
5 0

Answer:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}

And, the distance from P to (0, -2) will be:

\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}

So sum of the two distances must be 6. Therefore:

d_1+d_2=6

Now, by substitution:

(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6

Simplify. We can subtract the second term from the left:

\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}

Square both sides:

(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)

We can cancel the x² terms and continue squaring:

y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

-8y=36-12\sqrt{x^2+(y+2)^2}

We can divide both sides by -4:

2y=-9+3\sqrt{x^2+(y+2)^2}

Adding 9 to both sides yields:

2y+9=3\sqrt{x^2+(y+2)^2}

And, we will square both sides one final time.

4y^2+36y+81=9(x^2+(y^2+4y+4))

Distribute:

4y^2+36y+81=9x^2+9y^2+36y+36

The 36y will cancel. So:

4y^2+81=9x^2+9y^2+36

Subtracting 4y² and 36 from both sides yields:

9x^2+5y^2=45

And dividing both sides by 45 produces:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

LUCKY_DIMON [66]3 years ago
4 0

ANSWER:

We need to know a distance formula to find the locus.

By distance formula,

  • √[x² + (y - 2)²] + √[x² + (y + 2)²] = 6.
  • √(x² + y² - 4y + 4) + √(x² + y² + 4y + 4) = 6.

Let x² + y² + 4 = a.

  • √(a - 4y) + √(a + 4y) = 6.
  • √(a - 4y) = 6 - √(a + 4y).

Squaring both sides,

  • a - 4y = a + 4y + 36 - 12 × √(a + 4y).
  • 8y + 36 = 12 × √(a + 4y).
  • 2y + 9 = 3 × √(a + 4y).

Squaring again,

  • 4y² + 36y + 81 = 9a + 36y.
  • 9a - 4y² - 81 = 0.

Putting back the original value of a,

  • 9x² + 5y² - 45 = 0.

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Answer:

95% Confidence interval:  (0.0429,0.0791)      

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We are given the following in the question:

Sample size, n = 679

Number of anonymous websites, x = 42

\hat{p} = \dfrac{x}{n} = \dfrac{42}{679} = 0.0618

95% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

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tia_tia [17]

Answer:

8-6i

Step-by-step explanation:

Multiply out (3-i)^2 in order to find out which expression it is equivalent to.

1) Since the whole quantity is squared, write it out as (3-i)(3-i).

(3-i)^2\\(3-i)(3-i)

2) Multiply binomials by using the FOIL method. Multiply the terms that are listed first in each binomial, then the ones that are listed outermost when looking at both binomials, then innermost, and finally the last terms listed in each binomial. Simplify and combine like terms.

(3-i)(3-i)\\9 - 3i - 3i + i^2

9 - 6i + i^2

3)  i = \sqrt{-1}, so i ^2 must be (\sqrt{-1} )(\sqrt{-1}). Thus, i^2 is equivalent to -1. Knowing this, simplify and combine like terms.

9 - 6i + ((-\sqrt{1} )(\sqrt{-1} ))

9 - 6i -1\\8 - 6i

Thus, it is equivalent to  8 - 6i .

6 0
3 years ago
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